UVALive 2889（回文数字）

题意：给定i，输出第i个回文数字。

分析：1,2,3,4，……，9------------------------------------------------------------------------------------------9个

　　　11,12,13,14，……，19-------------------------------------------------------------------------------------9个

　　　101,111,121,131,141,151,161,171,181,191,202,212,222,232，……，979,989,999-------------------------------90个

　　　1001,1111,1221,1331,1441，……，9889,9999----------------------------------------------------------------90个

　　　10001,10101,10201,10301，……，99899,99999--------------------------------------------------------------900个

　　　规律是把回文串一分为二看，例如第四行，前两个数字是从10到99，共90个数字。而第三行也是从10到99，区别在于，需要去掉最后一位再反转，才是另一半（后两个数字）。

思路：先确定i所对应的回文数字的位数，再确定具体值。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cctype>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<iostream>
 7 #include<sstream>
 8 #include<iterator>
 9 #include<algorithm>
10 #include<string>
11 #include<vector>
12 #include<set>
13 #include<map>
14 #include<deque>
15 #include<queue>
16 #include<stack>
17 #include<list>
18 #define fin freopen("in.txt", "r", stdin)
19 #define fout freopen("out.txt", "w", stdout)
20 #define pr(x) cout << #x << " : " << x << "   "
21 #define prln(x) cout << #x << " : " << x << endl
22 typedef long long ll;
23 typedef unsigned long long llu;
24 const int INT_INF = 0x3f3f3f3f;
25 const int INT_M_INF = 0x7f7f7f7f;
26 const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
27 const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
28 const double pi = acos(-1.0);
29 const double EPS = 1e-6;
30 const int dx[] = {0, 0, -1, 1};
31 const int dy[] = {-1, 1, 0, 0};
32 const ll MOD = 1e9 + 7;
33 const int MAXN = 1000000 + 10;
34 const int MAXT = 10000 + 10;
35 using namespace std;
36 ll a[25];
37 void init()
38 {
39     ll tmp = ll(9);
40     for(int i = 1; i < 20; i += 2)
41     {
42         a[i] = a[i - 1] = tmp;
43         tmp *= ll(10);
44     }
45 }
46 ll POW(ll x)
47 {
48     ll w = 1;
49     for(ll i = 1; i <= x; ++i)
50         w *= ll(10);
51     return w;
52 }
53 int main()
54 {
55     init();
56     int n;
57     while(scanf("%d", &n) == 1 && n)
58     {
59         int cnt = 0;
60         while(n > a[cnt])
61         {
62             n -= a[cnt];
63             ++cnt;
64         }
65         if(cnt == 0)
66         {
67             printf("%d\n", n);
68             continue;
69         }
70         else if(cnt == 1)
71         {
72             printf("%d%d\n", n, n);
73             continue;
74         }
75         else
76         {
77             int tmp = cnt / 2;
78             char str[50];
79             memset(str, 0, sizeof str);
80             ll ans = POW(ll(cnt / 2)) + ll(n - 1);
81             sprintf(str, "%lld", ans);//把数字变为指定格式的字符串
82             printf("%s", str);
83             string s = string(str);
84             int len = s.size();
85             if(cnt % 2 == 0) s.resize(len - 1);
86             reverse(s.begin(), s.end());
87             printf("%s\n", s.c_str());
88         }
89     }
90     return 0;
91 }